Thay \(a=\dfrac{3}{5};b=\dfrac{2}{3}\) vào A, ta được:
\(A=10\left(\dfrac{3}{5}+\dfrac{2}{3}\right)-\dfrac{3}{5}\cdot\dfrac{2}{3}\)
\(=10\cdot\dfrac{9+10}{15}-\dfrac{2}{5}\)
\(=\dfrac{2}{3}\cdot19-\dfrac{2}{5}=\dfrac{38}{3}-\dfrac{2}{5}\)
\(=\dfrac{190-6}{15}=\dfrac{184}{15}\)